Friend's Question - B

1. Prove that: \( \cos\frac{2\pi^c}{7}+\cos\frac{4\pi^c}{7}+\cos\frac{6\pi^c}{7}=-\frac{1}{2} \)

Solution:

\begin{align} \text{LHS } & =\cos\frac{2\pi^c}{7}+\cos\frac{4\pi^c}{7}+\cos\frac{6\pi^c}{7} \\ & =2\cos\frac{\frac{2\pi^c}{7}+\frac{4\pi^c}{7}}{2}.\cos\frac{\frac{2\pi^c}{7}-\frac{4\pi^c}{4}}{2}+2\cos^2\frac{\pi^c}{7}-1 \\ & =2\cos\frac{3\pi^c}{7}\cos\frac{\pi^c}{7}+2\cos^2\frac{3\pi^c}{7}-1 \\ & =2\cos\frac{3\pi^c}{7}\left(\cos\frac{\pi^c}{7}+\cos\frac{3\pi^c}{7}\right)-1 \\ & =2\cos\frac{3\pi^c}{7}.2\cos\frac{\frac{\pi^c}{7}+\frac{3\pi^c}{7}}{2}.\cos\frac{\frac{\pi^c}{7}-\frac{3\pi^c}{7}}{2}-1 \\ & =4\cos\frac{3\pi^c}{7}\cos\frac{2\pi^c}{7}.\cos\frac{\pi^c}{7}-1 \\ & =\frac{2\left(2\sin\frac{\pi^c}{7}\cos\frac{\pi^c}{7}\right)\cos\frac{2\pi^c}{7}\cos\frac{3\pi^c}{7}}{\sin\frac{\pi^c}{7}}-1 \\ & =\frac{2\sin\frac{2\pi^c}{7}\cos\frac{2\pi^c}{7}\cos\frac{3\pi^c}{7}}{\sin\frac{\pi^c}{7}}-1 \\ & =\frac{\sin\frac{4\pi^c}{7}\cos\frac{3\pi^c}{7}}{\sin\frac{\pi^c}{7}}-1 \\ & =\frac{\sin\frac{4\pi^c}{7}\cos\left(\pi^c-\frac{4\pi^c}{7}\right)}{\sin\frac{\pi^c}{7}}\times\frac{2}{2}-1 \\ & =\frac{-2\sin\frac{4\pi^c}{7}\cos\frac{4\pi^c}{7}}{2\sin\frac{\pi^c}{7}}-1 \\ & =\frac{-\sin\frac{8\pi^c}{7}}{2\sin\frac{\pi^c}{7}}-1 \\ & =\frac{-\sin\left(\pi^c+\frac{\pi^c}{7}\right)}{2\sin\frac{\pi^c}{7}}-1 \\ & =\frac{\sin\frac{\pi^c}{7}}{2\sin\frac{\pi^c}{7}}-1 \\ & =\frac{1}{2}-1 \\ & =-\frac{1}{2} \\ & = \text{ RHS } \end{align}